Why is it necessary to travel 11km per second to escape the earth's gravity (escape velocity), when surely any speed will do, as long as you keep going up? | Notes and Queries | guardian.co.uk Why is it necessary to travel 11km per second to escape the earth's gravity (escape velocity), when surely any speed will do, as long as you keep going up? Luke, Wellington NZ If you start at less than 11 km/s at ground level and don't get any further 'push' (ie you slow down as you go up, like a stone thrown into the air) you will fall to the ground again. If you start at more than 11 km/s, you will never come down. Phil Cohen, Sydney Australia In a way, you are right. if you went straight up at any speed, eventually you would get so far from the earth that the effect of the Earth's gravity was negligable.

However, where the 11km per hour comes in is that it is the speed at which an object, rather than being held in orbit around the earth or falling back to it due to gravity, will be thrown outwards by sufficient centrifugal force to describe an open curve away from the planet. John Ramsey, Hackney UK Providing that you can keep your propulsion system going and the thrust is greater then gravitational pull back to Earth, yes any speed will do. Once you reach escape velocity however you are going fast enough to escape the Earth's gravitational pull without doing any more work, you can turn your engines off and you'll still keep going never to return. John Gresham, Waterloo Merseyside Escape velocity is calculated on the assumption that the only force applying is gravity, that is, only after the rocket engines (or whatever accelerated you to 11km/s) have stopped firing. Imagine yourself above the EarthÂs atmosphere travelling away from the Earth at high speed. EarthÂs gravitational pull is slowing you down, but as your distance from Earth increases, that pull weakens.

If you were initially travelling at less than the escape velocity then eventually EarthÂs pull would be enough to bring you to a stop and from then you on you would fall back to Earth. The escape velocity is the speed you need to avoid this fate and keep travelling forever, although you would need to travel at a much higher speed if you wanted to reach the stars in your lifetime. M Baillie, Sawston UK If you throw something up in the air, it will carry on rising until the force of gravity overcomes the velocity of the object and causes it to keep slowing until it stops and then starts falling back to earth. You are kind of correct in saying that you just need to "keep going up", but in order to keep going up you need a sufficient amount of velocity to overcome the force of gravity. The amount of velocity needed to escape our planets gravity is approx 11km per second, which is known as escape velocity. Steve, London UK The term Âescape velocityÂ refers to the initial velocity of an object (at the surface of the Earth). The assumption is that no further force is applied to the object once it has set off.

If the object starts at less than 11 km/sec, eventually the gravitational pull of the planet will pull it back to Earth before it escapes gravity. Of course any velocity will do if continued force is applied to the object, just as a car being driven uphill will keep going if enough power from the engine is applied via the accelerator. Jeremy Marchant, Stroud UK It isn't necessary. As you say, any speed will do. Escape velocity is simply the speed that theoretically would be sufficient to remove something from Earth's gravitational influence with no further impulse required. Michael Fisher, Brisbane Australia Escape velocity is the speed required to escape gravity in the absence of any force being applied. To keep going up requires the continual application of force. So yes, of course it is possible to keep going upwards and further from the earth, but without reaching escape velocity you would eventually be pulled back to earth when the fuel runs out. Clive Gordon, Ruislip UK Escape velocity is literally the velocity which an object would need to be projected in order to leave the surface of a planet or other body in space. If the object were subjected to a steady upward force just greater than its weight,instead of an explosion,it would eventually leave, but much more slowly.

It is interesting to note that the speed of air molecules is too low to allow them to escape the Earth, but greater than the Moon's escape velocity. That's why we have an atmosphere and the Moon hasn't. Is this rocket science? Alan Shepherd, Huddersfield UK You're right, but escape velocity is the speed a projectile (i.e. unpowered lump) would require to escape the Earth's gravity field. If you have enough power you can trundle up at your leisure. This point was ably covered in the old Peter Sellers movie "mouse on the Moon" which had a steam kettle spaceship. Rockets usually just have to get high enough to fall into orbit. However, you do need to get a rocket up to a high velocity before its fuel runs out if you want to send a probe to another planet. Craig Napier, Eagle Heights Australia It is not necessary to travel at 11km/s to escape the earth's gravity, any speed will do, as long as you keep going up.

11km/s is known as escape velocity because a body travelling up at this speed at the earth's surface will keep going up without any further force being applied; the reverse argument (that you must reach escape velocity to keep going up) does not follow, but anything travelling more slowly will need a bit of a push on the way up. Phil Barker, Edinburgh Scotland The escape velocity refers to an the speed an object needs to achieve to move from its current point in a gravitational field to infinity(as the gravitational field has no distinct end point), it is equal to the speed that the object would be traveling at if it was pulled by the same gravitational field from infinity to that same point. However this only applies for an unpowered mass that would be constantly decelerating due to the gravitational pull of the plant and you are quite right in saying a powered mass would be able to escape no matter what its speed. Steven, Glasgow You'd have to go up a very long way indeed to escape the earths gravity. Remember that this is what holds the moon in its orbit. An orbiting object has not escaped gravity, it is in constant freefall towards the earth. As long as it goes fast enough, it never reaches the ground because the earth curves away beneath it.

Howard, Haywards Heath West Sussex In theory, you're right, and if only Everest was taller, you could climb it all the way up to orbit as slowly as you damn pleased. Let's assume there's 2 ways into space, the slow way and the quick way. The slow way is to have some sort of propulsion engine, at least big enough to lift the vehicle's weight (e.g. a modern jet fighter can easily climb vertically, at speeds of 500MPH). The problem is that the jet engine can't get the oxygen it needs after about 5 miles up. The answer is to use a rocket motor, which carries all the fuel & oxygen it needs. But that turns out to be a lot of fuel (a HUGE lot of fuel) that you would be carrying, all the way up to 25,000 miles high or so where gravity becomes negligible and where, ideally, the last drop of fuel runs out. That brings us to the quick way: use that fuel in one quick burst, accelerate to 11km per second (25,000 MPH) as fast as possible, then coast the rest of the way with empty tanks and a dead engine.

That way, you're carrying spare fuel for the shortest possible time. Paul Reilly, London UK You are right, but in order to "keep going up" on sustained power you'd have to carry immense reserves of fuel to push you up. When Isaac Newton first described the concept of escape velocity he was thinking of a ball shot from a cannon. And modern rockets are much like that, their fuel tanks can only push them up for a few minutes and from then on they have rely on their own inertia. If they don't reach escape velocity by then, they fall back or remain in orbit. Alejandro Pareja, Madrid Spain True, in theory any speed would do. But in practice the difficulty comes from the trade-off between the rate at which you escape the Earth's gravity and the rate at which you consume the fuel needed to do so - and hence the fuel needed at take-off. If you ascended at 5mph, but had the typical fuel capacity of the shuttle and would be largely used up in 10 to 15 minutes, you'd only get up a couple of miles. Martin, Manchester UK Escape Velocity is the minimal velocity required to escape earth's gravity without applying any further force.

In this case, when any speed is maintained continuously a force is constantly applied. So the key point is the Force applied. Shravan Singh, Kolkata India An unpowered projectile needs escape velocity to travel arbitrarily high but a continually powered vehicle does not, of course. When the vehicle is in orbit above the atmosphere almost any force will do to push it further out - hence the ability of puny ion drives to push craft into higher orbits. It's getting into orbit that takes all the power. A grand space elevator is proposed to take the craft out to where the orbital speed is the rotation speed of the Earth's surface, so doing away with the flashy rocketry. Don't hold your breath. miles.felton, berkshire The 11 km/hr is the minimum speed necessary at the surface of the earth, to enable travelling any arbitrary distance from the earth without any means of propulsion. However, you are quite right that, if you have means of propulsion, such as a rocket ship, you can travel as slowly away from the surface as you like. Michael Hall, Canberra Australia A stone thrown upwards would need to achieve this speed, however the space shuttle coule go up as slow as required (Assuming enough fuel reserves).

The distinction is whether the flight is powered or not. Lee, Leeds UK There is a big difference between an object being aimed vertically upwards and shot out of a cannon and a powered rocket. The ball shot from the canon receives energy only as it passes through the barrel, from then on it is unpowered and slows down as it climbs through the earths gravitational field. Escape velocity refers to this case, not a powered rocket. Incidentally the mass of the ball does not effect the escape velocity if there is no atmospheric friction, which means that an elephant and a mouse would both have to be given the same escape velocity if launched from the surface of the moon! Chris Folland, Moshi Tanzania The term escape velocity presupposes that the object being considered is not subject to any external acceleration/force other than gravity. As such an object travels upwards it will of course be slowed by gravity, but at the same time an object that moves upwards from the earth the effect of gravity gradually dimishes. If you begin travelling upwards too slowly gravity will bring you back down to earth. If you start out travelling fast enough, whilst gravity will slow you down it will not be sufficient to bring you back down to earth. The escape velocity is the break point between these two alternatives.

The escape velocity is of course dependent upon the distance from the earth (or indeed any large body), diminshing as you travel away. Therefore if you started the earth's surface and travelled upwards at the escape velocity, although your velocity would diminish due to gravity it would still remain at what would be defined as the escape velocity. You could of course escape from earth's gravity if you could continuously move at even a very low speed. The problem is maintaining this speed against the pull of gravity. To do so you would need to introduce some other force, at which point the concept of escape velocity is no longer applicable. Paul Hoad, Sutton, UK At 11km/s you can successfully break orbit and escape the gravitational pull of the Earth. At 10km/s the Earth will eventually slow down your ascent till you begin falling back towards the ground. These values are at ground level. Once in orbit the escape velocity is lower than 11km/s. The greater the planet mass the greater the gravitational pull. To take off from the surface of MASSIVE Jupiter a rocket would need to be travelling at around 40 miles per second!! To escape the sun it is around 400 miles per second!

To escape a black hole you would have to travel beyond 300,000km/s - faster than the speed of light - which is impossible. That is why light cannot even escape from the surface! Paul Steele, Marske, UK Escape velocity is an energy thing. If you're going at escape velocity you don't need any more energy to escape from the Earth, because your kinetic energy is already enough (assuming you don't lose it; for example air drag). But you can go more slowly if you either spend more energy, or if you go ballistic from higher altitude. Ian Woollard, UK So how do you explain a helium filled balloon with no thrust and no persistent velocity, travelling at minimal speeds and still being able to leave Earth gravitational pull, assuming no gas is lost and the balloon remains in tact?

'Floating' into space would not require great thrust or huge amount of fuel but requires a helium to rocket ratio of at least 1:000000000001 recurring Thomas, Croydon England If we are talking about the planet of earth, the escape velocity is the speed primary needed to start a shell at the top, in this case must be the top speed of 11 km / s to end up outside the Earth's atmosphere, and therefore we can say that it is the speed of liberalization of gravity. Abdulah . I . Meftah, Farasan Islands, Saudi Arabia You're all wrong. The technology exists and has been used to escape earth's atmosphere at a much slower speed. However, the technology is highly classified and is only available to a select few. Drewey Whittaker, Albany Ga USA As per everyone (aside from the Area 51 theorist at the end), there's no reason why you can't provided your acceleration is >9.8 m/s then you will travel away from the Earth for as long as the fuel lasts.

One point nobody has made is that gravity is acceleration and to beat it you need opposite acceleration (something your chair is doing at them moment), so you would be accelerating away from earth. The closest Lagrange point (areas of flat space-time) of the moon would be the obvious target as you wouldn't then be pulled back to earth - because otherwise Earth's gravity will just take too long to reduce sufficiently - and anyway, you'd be stuck in deep space. As the fuel weighs a considerable amount, and needs sizable containers, there are efficiency reasons for burning it all as quickly as possible while dumping spent containers en route - standard rocket style. That said, a slow rocket would need to provide constant >10m/s thrust for the 300,000km or so it takes to get to your target L1. If your acceleration is a measly 12m/s (2.2m/s net acceleration), then even though L1 is 90% of the distance to the moon, by the halfway point you probably would have sufficient momentum to coast to L1. Even at 100km distant, gravity is only reduced by about 3% so it's not a major consideration.

However, your rocket wouldn't have to endure such huge tolerances, and I guess the structure could therefore be much lighter. Ok, now I might be making Mr Area 51 sound sensible but I'm beginning to see traction in a variant of the idea... Perhaps someone could correct me, but if a rocket accelerated directly up (not orbital acceleration as usual - I mean it's the same Newton requirement after all), burnt fuel reasonably quickly but never reaching anything like 25K, only enough to drift to a near stop at say L4, then couldn't we slingshot it on L4 (is it possible to slingshot a convex Lagrange? Surely it is.), slingshot the moon and then Earth, could escape velocity be reached more efficiently that way? Or say if you're going to the moon, why provide more energy than the journey requires by insisting on reaching EV with a orbital route? Am I missing something here? David, Peterborough, United Kingdom This is a big question - Can a Space vehicle take say three weeks to get to the Moon but travelling at a far slower speed. What is the fuel required to do that and would it be possible ?

The high velocity used at the moment is get into orbit around the earth however if you wanted to travel to the moon couldn't you do it slower but over a longer period of time ? Obviously if you didnt go into orbit around the moon you would have to be able to slow down so that you wouldn't crash into it. It does seem that if you had more time you could do things at slower pace Roy, Cape Town South Africa Speed Needed to Escape the Earth (Escape Velocity) Speed Needed to Escape the Earth (Escape Velocity) Standardized Result Buffa, Anthony J and Wilson, Jerry D. College Physics, Fourth Edition. Upper Saddle River: Prentice Hall, 2000. "Thus if a projectile or spacecraft could be given an initial upward speed of 11 km/s, it would leave the earth and not return." 11.0 km/s Mitton, Jacqueline. Cambridge Dictionary of Astronomy. New York: Cambridge University Press, 2001: 143. "The velocity of escape from the Earth's surface is about 11.2 km/s." 11.2 km/s Angelo, Joseph A. Jr. The Facts on File Space and Astronomy Handbook.

New York: Facts on File, 2002: 269. "Earth 11.2 km/s" 11.2 km/s Atkins, P.W. The Second Law. New York: Scientific American, 1984: 138. "The escape velocity is the speed an object must be given to escape from the Earth; it is 11.3 km/sec or 25,300 mph." 11.3 km/s Jones, A. Innovations in Astronomy. Santa Barbara: Helicon Publishing, 1999: 192. "In the case of the earth, the escape velocity is 11.2kps/6.9mps; the moon, 2.4kps/1.5mps; Mars, 5kps/3.1mps; and Jupiter, 59.6kps/37mps." 11.2 km/s Space travel is something we may now take for granted, but there are many factors we first have to look into. One thing we have to do is overcome the force of gravity holding us down.

We must be going at a very high speed, but how high? The escape velocity, as the minimum velocity that will allow a small body to escape from another body, can be calculated using the formula v = sqrt(2Gm/r), where G is the gravitational constant, r is the distance from the center of the body with a mass of m. This formula is derived using the idea of conservation of energy, where the sum of the kinetic energy and the gravitational energy of an object equal the energy at the maximum altitude. (+K) + (-U) [Earth] = K + U = 0 [at Infinity] Therefore What is Escape Velocity ? PhysLink.com. 21 May 2005. "From the surface of the Earth, escape velocity (ignoring air friction) is about 7 miles per second, or 25,000 miles per hour." 11.2 km/s Haven't we all wanted to escape the earth at one time or another in our lives? Well, to do that, you would have to go pretty quickly. The speed needed to escape the earth or "escape velocity" is about 11,200 meters per second or 7 miles per second.

Think about when you throw something up into the air. It has to come back because of the earth's gravitational pull, right? Well, yes, unless you throw it fast enough, which is highly unlikely no matter how fast you think your "fastball" is. Most commonly, the things that escape the earth are rockets fired by space agencies to explore space. With enough speed, a rocket can escape even the sun's gravity and go outside the solar system. The first artificial object to escape the earths gravity was in August of 1957. The object was a mineshaft cover sent into space by a nuclear device test explosion. It was sent out in Nevada. The earths rotation has a lot to do with the escape speed. This is why Nevada was a likely test cite rather than the North or South poles. When the object is closer to the equator, about 436 meters per second is added to the velocity. It doesn't seem like much when you look at the total speed, but trust me, it helps. The formula to find escape velocity is v = sqrt(2GM/r). M is the mass of the earth, G is the gravitational constant, r is the earths radius, and v is escape velocity.

Not as much to it as you thought, now is there? Mayya Kats -- 2005 How to Calculate Escape Velocity: 10 Steps - wikiHow How to Calculate Escape Velocity Two Parts: Understanding Escape Velocity Calculating Escape Velocity Community Q&A The escape velocity is the velocity necessary for an object to overcome the gravitational pull of the planet that object is on. [1] For example, a rocket going into space needs to reach the escape velocity in order to make it off Earth and get into space. Steps Understanding Escape Velocity 1 Define escape velocity. Escape velocity is the velocity of an object required to overcome the gravitational pull of the planet that object is on to escape into space. [2] A larger planet has more mass and requires a much greater escape velocity than a smaller planet with less mass. [3] 2 Begin with conservation of energy. Conservation of energy states that the total energy of an isolated system remains unchanged. In the derivation below, we will work with an Earth-rocket system and assume that this system is isolated.

In conservation of energy, we equate the initial and final potential and kinetic energies K {\displaystyle K_{1}+U_{1}=K_{2}+U_{2},} where Define kinetic and potential energy. Kinetic energy is energy of motion, and is equal to 1 {\displaystyle {\frac {1}{2}}mv^{2},} where is the mass of the rocket and v {\displaystyle v} is its velocity. Potential energy is energy that results from where an object is relative to the bodies in the system. In physics, we typically define the potential energy to be 0 at an infinite distance from Earth. Since the gravitational force is attractive, the potential energy of the rocket will always be negative (and smaller the closer it is to Earth). Potential energy in the Earth-rocket system is thus written as − is the mass of Earth, and r is the distance between the two masses' centers. 4 Substitute these expressions into conservation of energy. When the rocket achieves the minimum velocity required to escape Earth, it will eventually stop at an infinite distance from Earth, so K 0.

{\displaystyle K_{2}=0.} Then, the rocket will not feel Earth's gravitational pull and will never fall back to Earth, so U {\displaystyle {\frac {1}{2}}mv^{2}-{\frac {GMm}{r}}=0} 5 M r {\displaystyle {\begin{aligned}{\frac {1}{2}}mv^{2}&={\frac {GMm}{r}}\\v^{2}&={\frac {2GM}{r}}\\v&={\sqrt {\frac {2GM}{r}}}\end{aligned}}} v {\displaystyle v} in the above equation is the escape velocity of the rocket - the minimum velocity required to escape the gravitational pull of Earth. Note that the escape velocity is independent of the mass of the rocket m . {\displaystyle m.} The mass is reflected in both the potential energy provided by Earth's gravity as well as the kinetic energy provided by the movement of the rocket. Part 2 State the equation for escape velocity. v Gravity Applications Centripetal Force Newton's first law of motion says that an object's inertia will keep it from changing its speed and/or direction unless some force acts on it.

This means that satellites orbiting the Earth must be feeling some force that constantly deflects them toward the center of the Earth. If there was no force, they would travel in a straight line at a constant speed. If you whirl a ball attached to string around your head, it moves in a circular path around you because the string is always pulling the ball directly toward the hand grabbing the string. The ball wants to move in a straight line and the string is pulling it directly inward. The resulting deflection is a compromise: a circular path. The string is applying a centripetal force to the ball: an inward force. If you let go of the string, there is no centripetal force and the ball will fly off in a straight line because of its inertia. If you do not whirl the ball fast enough it will move inward to a smaller non-circular path around you. If you whirl the ball too fast, you may not be able to give it enough centripetal force to keep it in a circular path around you.

The amount of centripetal force needed to balance an object's inertia and keep it in a circular path of radius r is found from Newton's second law: the centripetal force = m v2 / r, where v and m are the object's speed and mass, respectively. The radius of the orbit r is the same as the distance between the moving object and the central body. Measuring Planet and Star Masses Now for orbits! Satellites are not being deflected by strings but by gravity. Gravity provides the centripetal force needed to keep the satellites in orbit. If you focus on the simple case of circular orbits, you can use the centripetal force formula above with the law of gravity to determine the mass of a planet or star. Simply set the force of gravity equal to the centripetal force and solve for the mass of the planet or star, M. (G M m) G . This assumes that the satellite's mass, m, is much less than the central object's mass so you can ignore the acceleration of the central object toward the satellite! How do you do that? Let's use this result to get an estimate of the mass of the Sun. You need to use something orbiting with a known radius and speed.

The Earth's orbit is roughly circular with radius = 1.5 × 1011 meters and the Earth moves with a speed 30,000 meters/second (= 30 km/s) in its orbit. The distance is given in meters to match the units of the speed. The distance unit of a meter is used because you will be using the gravitational constant G in your calculation and it uses the meter unit. When you do a calculation, you must be sure you check that your units match up or you will get nonsense answers. Plug the values into the mass relation: the Sun's mass = (30,000)2 × (1.5 × 1011)/(6.7 × 10-11) = 2 × 1030 kilograms. This is much larger than the Earth's mass so it was okay to ignore the Sun's movement toward the Earth. Using no approximations (ie., not assuming a circular orbit and including the Sun's motion toward the Earth) gives a value for the Sun's mass that is very close to this. Your answer does not depend on which planet you choose to use (here you used the Earth's orbit).

You would get the same value for the mass of the Sun if you had used any of the other planets orbital speeds and sizes. This relation tells you what you need to know in order to measure a planet's or star's mass: the orbital speed of a satellite and the distance it is from the center of the planet or star. Because the velocity is on top of the fraction, satellites are made to move faster if the mass of the central object is greater. At the same distance, a massive planet will exert more gravity force than a low-mass planet, so the massive planet will produce greater inward accelerations on satellites orbiting it. The satellites will, therefore, orbit at faster speeds. Sometimes the orbital period P is measured instead of the orbital velocity. The orbital period is the time it takes the object to travel the circumference of its orbit (for a circle, the circumference = 2pr, where p is approximately 3.1416). Recall that speed = (distance travelled)/time, so the speed v = the circumference of the orbit/orbital period.

When you substitute this for the speed in the mass relation above, you get M = Orbital Speed The mass formula above tells you that satellites orbiting massive planets must move faster than satellites orbiting low-mass planets at the same distance. Massive planets have stronger gravity than low-mass planets so a satellite orbiting a massive planet is accelerated by a greater amount than one going around a lesser mass planet at the same distance. To balance the stronger inward gravitational pull of the massive planet, the satellite must move faster in its orbit than if it was orbiting a lesser mass planet. Of course, this also applies to planets orbiting stars, stars orbiting other stars, etc. If you solve for the orbit speed, v, in the mass formula, you can find how fast something needs to move to balance the inward pull of gravity: v2 = (G M)/r . v = Sqrt[(G M)/r]. How do you do that? Find the orbital speed of Jupiter around the Sun. Jupiter's distance from the Sun is 5.2 A.U., or 7.8×1011 meters and the Sun's mass is 2×1030 kilograms.

The orbital speed of Jupiter around the Sun is Sqrt[6.7×10-11 × (2×1030)/(7.8×1011)] = Sqrt[1.718×108] = 1.3×104 meters/second, or 13 kilometers/second. What do you think you wouldd find if you used one of the Trojan asteroids, millions of times less massive than Jupiter, that orbits the Sun at 5.2 A.U.? Notice that the orbital speed v (and therefore, the orbital period P) does not depend on the mass of the satellite! The mass M in the formula above is the mass of the central object. If we consider the Sun-Jupiter-Trojan asteroid example in the "How do you do that?" box above (and recall the discussion in the Gravitational Acceleration section ), the Sun-Jupiter gravity force is MUCH greater than the Sun-Trojan asteroid gravity force but Jupiter has much greater inertia (resistance to change = mass).

The greater gravity force between the Sun and Jupiter is compensated for by the greater inertia of Jupiter so the acceleration of Jupiter will be the same as the acceleration of the Trojan asteroid---Jupiter and the Trojan asteroid will have the same orbital velocity and same orbital period. Escape Velocity If an object moves fast enough it can escape a massive object's gravity and not be drawn back toward the massive object. The critical speed needed to do this is the escape velocity. More specifically, this is the initial speed something needs to escape the object's gravity and assumes that there is no other force acting on the object besides gravity after the initial boost. Rockets leaving the Earth do not have the escape velocity at the beginning but the engines provide thrust for an extended period of time, so the rockets can eventually escape. The concept of escape velocity applies to anything gravitationally attracted to anything else (gas particles in planet atmospheres, comets orbiting the Sun, light trying to escape from black holes, galaxies orbiting each other, etc.). Using Newton's laws of motion and law of gravity, you can find that the escape velocity vesc looks very similar to the orbital speed: vesc = Sqrt[(2 G M)/r].

This is a factor Sqrt[2] larger than the circular orbital speed . Since the mass M is on top of the fraction, the escape velocity increases as the mass increases. More massive bodies exert greater gravity force, so escaping objects have to move faster to overcome the greater gravity. Also, the distance from the center of the object r is in the bottom of the fraction, so the escape velocity DEcreases as the distance increases. Gravity decreases with greater distance, so objects farther from a massive body do not need to move as quickly to escape it than those closer to it. How do you do that? Find the escape velocity from the surface of the Earth. Using the acceleration of gravity, you can find that the Earth has a mass of 6.0×1024 kilograms. The Earth's radius is 6.4×106 meters. Since the mass and distance from the center are in the standard units, you just need to plug their values into the escape velocity relation.

The Earth's surface escape velocity is Sqrt[2× (6.7×10-11) × (6.0×1024)/ (6.4×106)] = Sqrt[1.256×108] = 1.1×104 meters/second (= 11 km/s). Here are some other surface escape velocities: Moon = 2.4 km/s, Jupiter = 59.6 km/s, Sun = 618 km/s. Vocabulary newtonian mechanics - How long must escape velocity be maintained? - Physics Stack Exchange How long must escape velocity be maintained? 4 Escape velocity from Earth's surface is 11.2 Kilometres/second. How long would one need to maintain this escape velocity to actually escape Earth's gravitational pull? Must this 11.2 km/s velocity need to be maintained all the way up to geosynchronous orbit? up vote 17 down vote accepted No, the escape velocity doesn't need to be maintained for any length of time.

Escape velocity is the minimum speed you need to have at the Earth's surface to be able to escape the gravitational pull, without using a rocket or other continuous propulsion. In other words, ignoring all sources of gravity other than the Earth, if you launch a projectile straight up at escape velocity from Earth's surface, it will slow down as it rises but will never quite stop. Any slower, and the projectile will stop at some point and fall back to Earth. 2 Forgive me for being dense. If the escape velocity is maintained for only one second, surely the body will achieve an altitude of 11.2 kilometres in that period? At the end of that period, surely the gravity and air resistance will take over? – Everyone Jun 8 '12 at 21:22 8 Assume there's no air resistance - that's part of the definition of escape velocity. Given that, if escape velocity is maintained for one second, then yes, the body will achieve an altitude of 11.2 km in that time. Then, if its propulsion cuts out, it will be slowed down by gravity, but not enough to ever stop it.

– David Z♦ Jun 8 '12 at 21:26 4 To be more general, the escape velocity at any point is the instantaneous velocity needed then and there to get away. Note that escape velocity is really escape energy and is thus a scalar: any direction will do as long as it is a non-intersecting route. – dmckee♦ Jun 8 '12 at 23:39 This answer is interesting. You launch the ball straight up. About the slowing of the ball as it rises, will it be more important here than if I launch the ball horizontally ? – Nicolas Barbulesco Jun 7 '14 at 9:43 @Everyone — The escape velocity is from the Earth. Yes, after 1 s, the ball will have a speed < 11.2 km/s. But it will be far from the Earth. So it will be less subject to Earth's gravity. This is what I understand. And, as David said, there is no air resistance. We have decided so, because in physics we don't like air resistance.

^^ – Nicolas Barbulesco Jun 7 '14 at 9:50 Couldn't I escape Earth's gravity traveling only 1 mph? - Space Exploration Stack Exchange Couldn't I escape Earth's gravity traveling only 1 mph? up vote 91 down vote favorite 23 It is said that in order for an object or a projectile to leave Earth's gravitational pull, it must reach Earth's escape velocity, meaning reach a speed of 7 miles per second (~11 km per second). Well, as far as I understand, you could easily escape Earth's gravity even at 1 mph - directed away from the surface and you will eventually reach space. So why is the escape velocity 7 miles/s? Is it because the object has to gain a certain speed once it reaches orbit in order to maintain that altitude? Or is it because practically an object can't carry infinite amount of fuel, and so it has to reach a certain speed to maintain its orbit before all fuel is gone? Comments are not for extended discussion; this conversation has been moved to chat . – called2voyage♦ Aug 31 '16 at 17:22 up vote 86 down vote accepted The force of gravity decreases with distance.

It follows an inverse-square relationship... essential to know when you're grinding out the math, but not essential to a conceptual understanding. The fact that gravity decreases with distance means that at some distance, it will be negligible; an object sufficiently distant from Earth may be considered to have "escaped" Earth's gravity. In reality, the force of gravity has no distance limit; two objects would have to be at infinite distance from each other to have no gravitational interaction, but for practical purposes, one can think of finite distances where gravitational forces become small enough to ignore. Consider an object some large distance from Earth... right at the edge of what we would consider the Earth's gravitational "sphere of influence". Some tiny movement toward Earth will increase the gravitational attraction, accelerating the object toward Earth. The process will escalate with the object's velocity and acceleration increasing. If we ignore the effects of Earth's atmosphere, the object will continue its acceleration until it strikes the Earth's surface at some velocity. Now, let's reverse everything. The object magically launches up from Earth's surface at exactly the same speed as our falling object had at the instant of impact. As it rises up, gravity tugs on it and it slows down.

As it gets further away, gravity diminishes so it decelerates more slowly. Eventually, it gets to some distance where it has come to a stop, but Earth's gravity no longer has any effect on it. The velocity our object had at Earth's surface is Earth's escape velocity. In precise terms, a body's escape velocity is the velocity an object in "free fall" must have in order to escape the gravitational influence of that body - no more and no less. Technically, escape velocity can be specified for any distance from the center of a body, and the value will decrease with distance, but when a planet's escape velocity is stated, it is usually for the planet's surface. Mathematically, it is calculated as an integral of the body's gravitational acceleration from some specified distance to infinity. An object does not have to travel at escape velocity to escape a planet's gravity, but the same amount of energy needed to accelerate an object to escape velocity must be applied to an object (giving it potential energy) to lift it out of the planet's gravitational sphere of influence. The difference is that at escape velocity, the object needs no external influence to escape; at anything less than escape velocity, some external force must be applied.

Comments are not for extended discussion; this conversation has been moved to chat . – called2voyage♦ Aug 31 '16 at 17:22 up vote 64 down vote Escape velocity reduces as you get further away from the Earth. If you proceed upwards at a constant speed of 1 mph (which as noted will require continuous thrust to counteract gravity), you will eventually reach a distance where the escape velocity is equal to 1 mph. Then, you will have reached escape velocity and are no longer gravitationally bound to the Earth. This distance is extremely great; around 4×1012 km or 26000 AU. In practice, third-body effects (moon, sun, other planets) will dominate when you get beyond 105 km away from the Earth. So escape velocity is distance related. Thanks – user3339411 Jun 3 '14 at 23:08 2 By far the best, most direct answer to the question.

– Vedant Chandra Aug 18 '15 at 14:41 up vote 38 down vote To sum up the answers: the escape velocity is the velocity that, at a given distance, is sufficient to escape the gravitational field so that no additional energy (= acceleration) is needed. That is, if you are 26000 AU from Earth, you don't need any more fuel to counteract Earth's gravity, you just float away. However, when at Earth's surface, you will need additional acceleration to sustain the 1mph velocity - otherwise you just fall back down like the tossed ball. 4 Great answer - all the others are either flawed or far too complicated. – jwg Jun 5 '14 at 10:47 1 Practically speaking, there are other objects in the vicinity that will affect you so this is just an academic exercise. – Brian Jun 6 '14 at 11:58 Why 'just'? By that standard, isn't every orbital calculation an academic exercise? What application of Newtonian physics would not be?

– Spike0xff Jun 8 '14 at 1:01 up vote 23 down vote You are confusing velocity and acceleration. If you were to jump standing on the surface of the Earth you might experience 8 m/s which is 17 mph velocity upward, but the acceleration of gravity would act to retard your motion, slowing your velocity down. If you have a high enough velocity, the effect of (de) acceleration can not slow you down before you get far enough away from the gravitational source. So if you could keep a constant velocity of 1 mph, you would defiantly be able to escape the earth. The problem is that would require constant thrust. If you're going 11 km/s then you can just relax and watch the world shrink in your rear view mirror. 2 Also note that the amount of energy needed to maintain that upward velocity of 1 mph to the point where escape velocity is 1 mph vastly exceeds the amount of energy needed at attain a velocity of 11 km/s right off the bat. – David Hammen Jun 3 '14 at 20:50 10 I'm pretty sure defiantly is a typo for definitely, but I really like it in this case.

Defiantly escaping the Earth! Who needs it anyway? – Tim Gostony Jun 3 '14 at 22:54 4 @TimGostony, you're right it was a typo, but I'm now defiantly not changing it! :D – ThePlanMan Jun 3 '14 at 22:55 1 +1 just for defiantly accepting the DYAC. – David Hammen Jun 4 '14 at 3:58 2 @DavidHammen that is actually incorrect. The amounts of energy would be equal. The thrust to maintain 1mph would only have to counter the potential energy gained by the increased height. The KE of a 1kg object at 11.115 km/s is 62.217 MJ. The potential energy of a 1kg object gained rising from 6.4x10^6 m to 4x10^15 m is 62.218 MJ, and I am going to count the difference from rounding. – Godric Seer Jun 6 '14 at 16:05 Thanks. Yes, this makes sense.

– user3339411 Jun 3 '14 at 23:03 One of the best answers to the question, pointing out that the escape speed makes sense only for an unpowered spacecraft which must counter the gravity until it is insignificant. – mins Feb 7 '15 at 20:02 up vote 7 down vote The key difference is that "escape velocity" is how fast you would have to throw a stone straight up from the Earth's surface (ignoring air drag), for it to escape from Earth's gravitational influence. It would be coasting the whole way, always losing speed due to Earth's gravitational pull. If, on the other hand, you have a rocket engine with sufficient fuel, you can just keep rising slowly (1 mph), which is almost a hover, until you've gotten way out into space and Earth's gravity is overwhelmed by the Sun, Jupiter, etc. You could keep throttling back to maintain the same upward speed (gravity decreases with distance, and the rocket carries less fuel) if you wanted to, or let the rocket speed up.

up vote 6 down vote Unless you are very far away from Earth, if you are only moving away at 1 mph the gravity of Earth will pull you back to Earth (assuming you do not have an infinite fuel supply to maintain a 1mph thrust). So you are correct when you say Is it because the object has to gain a certain speed once it reaches orbit in order to maintain that altitude. Think of a ball tossed in the air, it starts by moving quickly, but as it rises higher it goes slower, than stops and falls back down. At some point it is moving away from Earth at 1mph, but gravity overcomes that momentum. Air Resistance has some impact on the ball, but you can throw horizontally much farther than you can up. Gravity works pretty much them same on the surface of the Earth as it does a 1000 miles up. When you throw something horizontally it falls towards the earth in an arc, attracted by the gravity of the Earth. If it is moving fast enough the curvature of the Earth will match the arc of the falling object, this is called Orbital speed and the object will not hit the earth. up vote 4 down vote Looking at this in another way, consider the concept of gravity wells .

The gravity well of course is not a "real", physical well, but it is a commonly used metaphor to describe how much energy is required to escape from the gravitational effect of a body, and it provides a reasonably straight-forward way of answering your question. (Space buffs, bear with me below; this is meant as an explanation, not a university-level physics and astronomy lecture.) If you are at or near the bottom of a gravity well (say, at the surface of the Earth) and want to climb out of that well, you basically have two options. Either climb very fast for a short distance (this is the approach taken for getting off the surface of the Earth, for reasons stated in other answers), or climb slowly for a much longer distance (this works once you are far enough away from the body forming the gravity well that the predominant gravitational forces acting on you are small or negligible). Each way of looking at it represents the same thing: you provide some sort of energy input, usually in terms of fuel of some kind, which is used to climb the "side" of the gravity well.

The energy provided as input becomes potential energy as you climb farther from the surface, and at some point, your potential energy exceeds the gravitational pull at that point of the body that forms the gravity well; you "continue on a tangent" and move straight on from that point forward rather than following the curve of the gravity well. Once that happens, you have reached escape velocity from that body. If you don't climb far enough for your rate of climb at the time you stop actively climbing, then when you stop climbing (let's assume you cannot grab hold of anything, because in space there is nothing to hold on to) you will fall back toward the body that forms the gravity well you are trying to climb out of; you did not attain escape velocity. Of course, there are usually multiple gravitational forces to contend with at any one point. However, one of them will project a stronger force on you than the others; that's the concept behind the sphere of influence . Near Earth (yes, that most definitely includes low Earth orbit), it's Earth's gravity that dominates; take a trip to Luna and its gravity will exert the greater force once you pass the Earth-Moon system L1 Lagrangian point .

The "depth" of a gravity well is often given as its escape velocity, in km/s or some other convenient measurement of velocity, taken at the bottom of the well. Hence, the depth of Earth's gravity well is approximately 11.2 km/s, which is the escape velocity at the Earth's surface. Wikipedia gives the escape velocity at 9,000 km above the Earth's surface as 7.1 km/s , but as we have seen in other answers, getting to 9,000 km above the surface itself takes a lot of energy, negating the gain from the lower "absolute" speed necessary to break free of the Earth's gravity. What is escape velocity? What is escape velocity? 1 2 Next 0 Escape velocity is the speed a moving object must have in order to escape from the gravitational pull exerted by another object. For example, the escape velocity from Earth’s surface is equal to about 6.8 miles per second (11 km/s). Anything that wants to escape Earth’s gravitational pull must go at least 6.8 miles per second (11 km/s), no matter what the thing is – a rocket ship or a baseball.

The escape velocity of an object depends on how compact it is; that is, on the ratio of its mass to radius. A black hole is an object so compact that, within a certain distance of it, even the speed of light is not fast enough to escape. more hubblesite.org Is this page helpful to you? Your anonymous feedback will help us improve. Related Videos 0 Escape velocity is the velocity at which a body must travel before it breaks free of the gravitational pull of a celestial body. It varies widely based on the body's mass. The Earth's escape velocity is 11.186 km/s (25,022 mph or about Mach 37), which can only reached by powerful booster rockets. By comparison, the Space Shuttle's top speed is about 7.6 km/sec, enough to reach Low Earth Orbit but not escape the Earth's gravitational pull entirely. The escape velocity on Mars' moon Phobos is about 11 m/s, or 25 mph. The body's mean diameter is 11 km (6.8 mi).

Its smaller cousin, Demos, with a mean diameter of 6 km (3.7 mi), has an escape velocity of only 6.9 m/s (15 mph), meaning that you could probably jump right off it if you tried hard enough. The gravitational force decreases with the square of the distance, meaning that if a certain distance away from the body the gravity is only half that on the surface, at twice that distance the gravity will be four times less, eight times ... more wisegeek.com Write a comment... 0 "Escape velocity" is a misnomer, because it doesn't refer to a velocity, but rather a speed. It is properly "escape speed". The escape speed for a mass is the speed required to travel an infinite distance from said object. In other words (and taking Re to be the radius of the Earth), escape speed for Earth is the speed at r = Re such that speed at infinite r is 0. Obviously, that would take an infinite amount of time, but it is a useful theoretical exercise. As speed at infinite r is 0, kinetic energy is also 0.

Gravitational potential energy is defined to be 0 at infinite distance, so at infinite distance, the total energy of the object is 0. Due to conservation of energy, that must be the energy of the object on the surface of the Earth. Therefore: E = (1/2)*m*v ^ 2 - (G*M*m)/Re = 0, when E is total energy, m is the mass of the object, v is the escape speed, G is the gravitational constant, M is the mass of the Earth and Re is as above. Solving for v, we find: v = sqrt( (2*G*M) / ... answerbag.com

Question: What name is given to the speed it is necessary for a rocket to achieve in order for it to overcome the Earth's gravitational pull?
Answer: ESCAPE VELOCITY